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%TCIDATA{Created=Tue Jun 08 12:10:31 2004}
%TCIDATA{LastRevised=Wednesday, June 11, 2008 09:36:37}
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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\begin{document}
El dominio de la funci\'{o}n $\ y=\dfrac{x-9}{\left(  x-3\right)  ^{2}\left(
x+1\right)  }$ es:\newline\qquad a) $\left\{  x\in\Re:x\neq-1\text{ \ }y\text{
\ }x\neq3\right\}  \qquad\qquad\qquad\ $b) $\left\{  x\in\Re:x\neq1\text{
\ }y\text{ \ }x\neq3\right\}  \medskip$\newline\qquad c) $\left\{  x\in
\Re:x\neq-1,\text{ \ }x\neq3\text{ \ }y\text{ \ }x\neq9\right\}  \qquad$d)
$\left\{  x\in\Re:x=-1\text{ \ }y\text{ \ }x=3\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x-2}{\left(  x-1\right)  ^{2}\left(
x+1\right)  }$ es:\newline\qquad a) $\left\{  x\in\Re:x\neq1\text{ \ }y\text{
\ }x\neq-1\right\}  \qquad\qquad\qquad\ $b) $\left\{  x\in\Re:x\neq1\text{
\ }y\text{ \ }x\neq2\right\}  \medskip$\newline\qquad c) $\left\{  x\in
\Re:x\neq-1,\text{ \ }x\neq2\text{ \ }y\text{ \ }x\neq-1\right\}  \qquad$d)
$\left\{  x\in\Re:x=-1\text{ \ }y\text{ \ }x=1\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{5x-2}{\left(  3x-1\right)
^{2}\left(  x+4\right)  }$ es:\newline a) $\left\{  x\in\Re:x\neq\dfrac{1}%
{3}\text{ \ }y\text{ \ }x\neq-4\right\}  \qquad\qquad\qquad\ $b) $\left\{
x\in\Re:x\neq-\dfrac{1}{3}\text{ \ }y\text{ \ }x\neq4\right\}  \medskip
$\newline c) $\left\{  x\in\Re:x\neq\dfrac{1}{3},\text{ \ }x\neq-4\text{
\ }y\text{ \ }x\neq\dfrac{2}{5}\right\}  \qquad$d) $\left\{  x\in
\Re:x=-3\text{ \ }y\text{ \ }x=4\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x^{2}-16}{\left(  x-4\right)
^{2}\left(  x+4\right)  }$ es:\newline\qquad a) $\left\{  x\in\Re
:x\neq-4\text{ \ }y\text{ \ }x\neq4\right\}  \qquad\qquad\qquad\ $b) $\left\{
x\in\Re:x\neq4\text{ \ }\right\}  \medskip$\newline\qquad c) $\left\{  x\in
\Re:x\neq-4\right\}  \qquad\qquad\qquad\ \ $d) $\left\{  x\in\Re:x=-4\text{
\ }y\text{ \ }x=4\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x-1}{\left(  x+6\right)  ^{2}\left(
x+2\right)  }$ es:\newline\qquad a) $\left\{  x\in\Re:x\neq-6\text{ }y\text{
\ }x\neq-2\text{\ }\right\}  \qquad\qquad\ $b) $\left\{  x\in\Re:x\neq6\text{
\ }y\text{ \ }x\neq2\right\}  \medskip$\newline\qquad c) $\left\{  x\in
\Re:x\neq1\right\}  \qquad\qquad\qquad\qquad\qquad\ $d) $\left\{  x\in
\Re:x=1\text{ \ }y\text{ \ }x=6\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{3x-2}{\left(  x^{2}-9\right)  \left(
x+1\right)  }$ es: \newline\qquad\medskip a) $\left\{  x\in\Re:x\neq-1\text{
},\text{\ }x\neq3\text{ \ }y\text{ \ }x\neq-3\right\}  $\qquad b) $\left\{
x\in\Re:x\neq1\text{ },\text{\ }x\neq3\text{ \ }y\text{ \ }x\neq-3\right\}
\ $\newline\qquad c) $\left\{  x\in\Re:x\neq1\text{ \ }y\text{ \ }%
x\neq3\right\}  $\qquad d) $\left\{  x\in\Re:x=-1\text{ \ }y\text{
\ }x=3\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x-1}{\left(  x+2\right)  ^{2}\left(
x^{2}-16\right)  }$ es: \newline\qquad\medskip a) $\left\{  x\in\Re
:x\neq-2\text{ },\text{\ }x\neq4\text{ \ }y\text{ \ }x\neq-4\right\}  $\qquad
b) $\left\{  x\in\Re:x\neq-2\text{ },\text{\ }x\neq4\text{ \ }y\text{ \ }%
x\neq1\right\}  \ $\newline\qquad c) $\left\{  x\in\Re:x\neq1\text{
},\text{\ }x\neq4\text{ \ }y\text{ \ }x\neq-4\right\}  $\qquad d) $\left\{
x\in\Re:x\neq2\text{ },\text{\ }x\neq4\text{ \ }y\text{ \ }x\neq1\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x+1}{\left(  x-3\right)  ^{2}\left(
x-1\right)  }$ es: \newline\qquad\medskip a) $\left\{  x\in\Re:\text{\ }%
x\neq3\text{ \ }y\text{ \ }x\neq1\right\}  $\qquad b) $\left\{  x\in\Re
:x\neq-1,\text{ \ }x\neq3\text{ \ }y\text{ \ }x\neq1\right\}  \ $%
\newline\qquad c) $\left\{  x\in\Re:x\neq-3,\text{ \ }x\neq3\text{ \ }y\text{
\ }x\neq1\right\}  $\qquad d) $\left\{  x\in\Re:x=-1\text{ \ }y\text{
\ }x=3\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x+1}{\left(  x^{2}-9\right)  \left(
x-1\right)  ^{2}}$ es: \newline\qquad\medskip a) $\left\{  x\in\Re
:x\neq-3,\text{ \ }x\neq3\text{ \ }y\text{ \ }x\neq1\right\}  $\qquad b)
$\left\{  x\in\Re:x\neq-1\text{ \ }y\text{ \ }x\neq1\right\}  \ $%
\newline\qquad c) $\left\{  x\in\Re:x\neq1\text{ \ }y\text{ \ }x\neq3\right\}
$\qquad d) $\left\{  x\in\Re:x\neq-1,\text{ \ }x\neq3\text{ \ }y\text{
\ }x\neq1\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{2x+5}{\left(  x-2\right)  ^{2}\left(
x^{2}-1\right)  }$ es: \newline\qquad\medskip a) $\left\{  x\in\Re
:x\neq-1,\text{ \ }x\neq2\text{ \ }y\text{ \ }x\neq1\right\}  $\qquad b)
$\left\{  x\in\Re:\text{ }x\neq2\text{ \ }y\text{ \ }x\neq-1\right\}
\ $\newline\qquad c) $\left\{  x\in\Re:x\neq-1\text{ \ }y\text{ \ }%
x\neq1\right\}  $\qquad d) $\left\{  x\in\Re:\text{ }x\neq2\text{ \ }y\text{
\ }x\neq1\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x-7}{\left(  x-2\right)  \left(
x+3\right)  (x^{2}-9)}$ es: \newline\qquad\medskip a) $\left\{  x\in\Re
:x\neq-3,\text{ \ }x\neq3\text{ \ }y\text{ \ }x\neq2\right\}  $\qquad b)
$\left\{  x\in\Re:x\neq7,\text{ }x\neq-3,\text{ \ }x\neq3\text{ \ }y\text{
\ }x\neq2\right\}  \ $\newline\qquad c) $\left\{  x\in\Re:x\neq-3,\text{
\ }x\neq7\text{ \ }y\text{ \ }x\neq2\right\}  $\qquad d) $\left\{  x\in
\Re:x\neq-3\text{ \ }y\text{ \ }x\neq2\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x-9}{\left(  x-2\right)  ^{2}\left(
x-1\right)  (x^{2}-36)}$ es: \newline\qquad\medskip a) $\left\{  x\in\Re
:x\neq1,\text{ \ }x\neq2,\text{ }x\neq6\text{\ \ }y\text{ \ }x\neq-6\right\}
$\qquad b) $\left\{  x\in\Re:x\neq1,\text{ \ }x\neq2\text{\ \ }y\text{
\ }x\neq36\right\}  \ $\newline\qquad c) $\left\{  x\in\Re:x\neq1,\text{
\ }x\neq2\text{\ \ }y\text{ \ }x\neq6\right\}  $\qquad d) $\left\{  x\in
\Re:x\neq9,\text{ }x\neq1,\text{ \ }x\neq2,\text{ }x\neq6\text{\ \ }y\text{
\ }x\neq-6\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x-8}{\left(  x-1\right)  ^{2}\left(
x-3\right)  }$ es: \newline\qquad\medskip a) $\left\{  x\in\Re:x\neq1\text{
\ }y\text{ \ }x\neq3\right\}  $\qquad b) $\left\{  x\in\Re:x\neq-1\text{
\ }y\text{ \ }x\neq3\right\}  \ $\newline\qquad c) $\left\{  x\in\Re
:x\neq-1,\text{ }x\neq1\text{ \ }y\text{ \ }x\neq3\right\}  $\qquad d)
$\left\{  x\in\Re:x\neq-1,\text{ \ }x\neq3\text{ \ }y\text{ \ }x\neq8\right\}
$

El dominio de la funci\'{o}n $\ y=\dfrac{x-5}{\left(  x-3\right)  ^{2}\left(
x^{2}-4\right)  }$ es: \newline\qquad\medskip a) $\left\{  x\in\Re
:x\neq-2,\text{ \ }x\neq2\text{ \ }y\text{ \ }x\neq3\right\}  $\qquad b)
$\left\{  x\in\Re:x\neq-3,\text{ \ }x\neq3\text{ \ }y\text{ \ }x\neq2\right\}
\ $\newline\qquad c) $\left\{  x\in\Re:x\neq5,\text{ }x\neq-2,\text{ \ }%
x\neq2\text{ \ }y\text{ \ }x\neq3\right\}  $\qquad d) $\left\{  x\in\Re
:x\neq3,\text{ \ }x\neq2\text{ \ }y\text{ \ }x\neq5\right\}  $

El dominio de la funci\'{o}n $\ y=\dfrac{x+1}{\left(  x-2\right)  ^{2}\left(
x-1\right)  (x^{2}-25)}$ es: \newline\qquad\medskip a) $\left\{  x\in\Re
:x\neq5,\text{ }x\neq-5,\text{ \ }x\neq2\text{ \ }y\text{ \ }x\neq1\right\}
$\qquad b) $\left\{  x\in\Re:x\neq5,\text{ }x\neq-5,\text{ \ }x\neq1,\text{
}x\neq-1\text{ \ }y\text{ \ }x\neq2\right\}  \ $\newline\qquad c) $\left\{
x\in\Re:x\neq5,\text{ }x\neq2,\text{ \ }x\neq1\text{ \ }y\text{ \ }%
x\neq-1\right\}  $\qquad d) $\left\{  x\in\Re:x\neq5,\text{ }x\neq
2\text{\ \ }y\text{ \ }x\neq1\right\}  $


\end{document}